山东省青岛市2022-2023学年高二数学上学期期中考试试题（Word版附答案）

这是一份山东省青岛市2022-2023学年高二数学上学期期中考试试题（Word版附答案），共8页。试卷主要包含了已知一组数据，若数列是等比数列，则的值是等内容，欢迎下载使用。
2022-2023学年度第一学期期中学业水平检测  高二数学试题 本试卷4页，22小题，满分150分．考试用时120分钟． 注意事项：1．答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将准考证号条形码粘贴在答题卡上的指定位置。2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需要改动，用橡皮擦干净后，再选涂其它答案标号。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。3．考试结束后，将本试卷和答题卡一并交回。 一、单项选择题：本大题共8小题，每小题5分，共40分。在每小题给出的四个选项中，只有一项是符合题目要求的。   1．某校把纸笔测试、实践能力、成长记录三项成绩分别按、、的比例记入学期总评成绩，分以上为优秀，甲、乙、丙三人的各项成绩如下表(单位：分)： 纸笔测试实践能力成长记录甲乙丙则学期总评优秀的是 A．甲 B．乙、丙 C．甲、乙 D．甲、丙2．数列的一个通项公式是 A． B． C． D．3．某社会调查机构就某地居民的月收入情况调查了人，并根据所得数据画了样本的频率分布直方图（如下图）．为了分析居民的月收入与年龄、学历、职业等方面的关系，要从这人中再用分层抽样方法抽出人作进一步调查，则在（元）月收入段应抽出人数是 A． B． C． D．        4．已知一组数据：，则其第百分位数为 A． B． C． D．   5．从一批产品（其中正品、次品都多于件）中任取件，观察正品件数和次品件数，则下列事件是互斥事件的是：①恰有一件次品和恰有两件次品               ②至少有一件次品和全是次品③至少有一件正品和至少有一件次品           ④至少有一件次品和全是正品 A．①② B．①④ C．③④ D．①③   6．若数列是等比数列，则的值是 A． B． C． D．   7．有个相同的小球，分别标有数字，从中有放回地随机取两次，每次取个球，甲表示事件“第一次取出的球的数字为”，乙表示事件“第二次取出的球的数字为”，丙表示事件“两次取出的球的数字之和为”，丁表示事件“两次取出的球的数字之和为”，则 A．丙与丁相互独立  B．甲与丙相互独立 C．乙与丙相互独立  D．乙与丁相互独立   8．集合论是德国数学家康托尔于十九世纪末创立的，希尔伯特赞誉其为“数学思想的惊人产物，在纯粹理性范畴中人类活动的最美表现之一”．取一条长度为的线段，将它三等分，去掉中间一段，留下的两段分割三等分，各去掉中间一段，留下更短的四段，……，将这样操作一直继续下去，直至无穷．由于在不断分割舍弃过程中，所形成的线段的数目越来越多，长度越来越小，在极限情况下，得到一个离散的点集，称为康托尔三分集．若在前次操作中共去掉的线段长度之和不小于，则的最小值为         （参考数据：） A． B． C． D．   二、多项选择题：本大题共4小题，每小题5分，共20分。在每小题给出的四个选项中，有多项符合题目要求。全部选对的得5分，部分选对的得2分，有选错的得0分。 9．对划艇运动员甲、乙二人在相同的条件下进行了次测试，测得他们最大速度的数据如下，甲：；乙：．根据以上数据，则A．他们最大速度的平均值相等  B．他们最大速度的中位数相等C．同样情况下，甲运动员的发挥比乙更稳定 D．同样情况下，乙运动员的发挥比甲更稳定10．已知数列是公比的正项等比数列， 是与的等比中项 ，是与等差中项，则A． B． C． D．11．如图，由到的电路中有个元件，分别标为元件，元件，元件，元件，电流能通过元件，元件的概率都是，电流能通过元件，元件的概率都是，电流能否通过各元件相互独立．已知元件，元件中至少有一个能通过电流的概率为，则    A．  B．元件和元件恰有一个能通的概率为C．元件和元件都通的概率是 D．电流能在与之间通过的概率为12．已知数列满足：，则A．  B．对任意恒成立C．不存在正整数使成等差数列 D．数列为等差数列  三、填空题：本大题共4个小题，每小题5分，共20分。 13．对某种新品电子元件进行寿命终极度实验，统计情况如下：寿命（h）个数估计优质品（寿命h以上者）的概率为             ． 14．已知从某班学生中任选两人参加农场劳动，选中两人都是男生的概率是，选中两人都是女生的概率是，则选中两人中恰有一人是女生的概率为             ． 15．已知等差数列的公差为，且是等比数列的前三项，则数列的前项和             .16．已知某区甲、乙、丙三所学科基地学校的数学强基小组人数分别为人，在一次统一考试中，该区三所学校强基学生的平均分分别为，方差分别为，则该区所有数学强基学生成绩的平均数             ，方差             ．（第一空分；第二空分）  四、解答题：共70分。解答应写出文字说明，证明过程或演算步骤。 17．（10分）袋子中放有大小和形状相同的小球若干个，其中红色小球1个，黄色小球个，蓝色小球个，从袋子中随机抽取个小球，设取到蓝色小球为事件，且事件发生的概率是．（1）求的值；（2）从袋子中不放回地随机抽取个小球，若每次取到红色小球得分，取到黄色小球得分，取到蓝色小球得分，设第一次取出小球后得分为，第二次取出小球后得分为，记事件为“”，求事件发生的概率．     18．（12分）某学校高一级部根据同年龄段女生的身高数据绘制了频率分布直方图，其中身高的变化范围是 (单位：厘米)，样本数据分组为，，，，，．（1）求值；（2）已知样本中身高大于厘米的人数是，求出样本总量的数值和身高超过厘米的人数；（3）求样本中位数的值．   19．（12分）已知是数列的前项和，且．（1）求的通项公式；（2）若，求． 20．（12分）某区三所学校有意愿报考名校自招的人数分别为人，受疫情因素影响，该区用分层随机抽样的方法从三所学校中抽取了名学生，参加了该区统一举办的现场小范围自招推介说明会．（1）从这名中随机抽取名学生进行座谈和学情调查，求这名学生来自不同学校的概率；（2）若考生小张根据自身实际，报考了甲乙两所名校的自招，设通过甲校自招资格审核的概率为，通过乙校自招资格审核的概率为，已知通过两所学校自招资格审核与否是相互独立的，求小张至少能通过一所学校自招资格审核的概率． 21．（12分）已知为数列的前项和，，为数列的前项和，．（1）求数列的通项公式；（2）若对任意恒成立，求正实数的取值范围． 22．（12分）已知数列满足：．（1）证明：数列为等比数列，并求数列的通项公式；（2）证明：；（3）若正整数，，记．（ⅰ）求；（ⅱ）证明：．   2022-2023学年度第一学期期中学业水平检测高二数学评分标准    一、单项选择题：本大题共8小题，每小题5分，共40分。  1-8： CBBC  BCDA二、多项选择题：本大题共4小题，每小题5分，共20分。9．AD       10． BC      11．ACD      12．ABD三、填空题：本大题共4个小题，每小题5分，共20分。13．   14．   15．   16．（1）；（2）（第一空分；第二空分）四、解答题：共70分。解答应写出文字说明，证明过程或演算步骤。 17．（10分）解：（1）由题意，从袋子中随机抽取个小球，共有个结果······················2分每个结果可能性相同，其中事件发生有种结果································3分所以，解得··························································4分(2)把红色小球记为；黄色的小球记为；蓝色小球记为；则两次不放回地取出小球的组合情况可用表格表示为   共个样本点，························································6分其中事件包含的样本点有,,,，共个，·······································8分所以······························································10分 18．（12分）解：（1）由频率分布直方图的性质···································································2分解得·······························································3分（2）身高大于厘米的样本的频率是········································4分所以样本总量························································5分身高超过厘米的频率为·················································6分所以身高超过厘米的人数；··············································7分（3）因为身高位于的频率为·············································8分身高位于的频率为·····················································9分所以中位数应该······················································10分由································································11分解得······························································12分19．（12分）解：（1）由题知：时，·······························································1分时，（）····························································2分 也符合（）式························································3分所以·······························································4分（2）因为，所以时，；时，·············································6分时，························································8分时， ·······················································11分综上：····························································12分 20．（12分）解：（1）用分层随机抽样的方法从三个学校中一共抽取了名选手参加全市集训，现三所学校应该抽取的人数分别为··········································1分 设来自学校的三名学生分别为;来自学校的学生为；来自学校的两名学生分别为从这名中随机抽取名学生进行座谈和学情调查，样本空间 共包含 个样本点······················································3分记这名学生来自不同学校为事件，事件含共 个样本点，························································5分所以·······························································6分（2）记小张至少能通过一所学校自招资格审核为事件，通过甲学校自招资格审核为事件，通过乙学校自招资格审核为事件，则事件“至少通过一所学校自招资格审核”的对立事件是“两所学校都通不过”，因为与相互独立，所以与相互独立········································8分所以 答：小张至少能通过一所学校自招资格审核的概率为··························12分  21．（12分）解：由题意，对任意，有①当时，，可得，，所以·················································1分当时， ②①②得：·····························································3分所以，即····························································4分所以，对任意，数列是以为首项，以为公比的等比数列·························5分所以·······························································6分（2）因为···························································7分所以·······························································8分所以·······························································9分可以看出，随着的增大而增大，所以，且对任意，····················································10分所以恒成立，有，······························································11分所以，所以······························································12分 22．（12分）解:（1）因为，所以···················································1分又因为·····························································2分所以是以为首项，为公比的等比数列·······································3分所以所以·······························································4分（2）因为···························································6分所以·······························································7分（3）（ⅰ）由题知：··················································8分又因为·····························································9分所以······························································10分（ⅱ）因为又因为·················································11分所以 12分